Equation with Two Exponential Terms

Date: 06/27/2009 at 15:34:36
From: Dan
Subject: find ordered pair (a,b) for 3^a + 7^b giving a perfect square
Any perfect square can be expressed as n^2, find all ordered pairs
for 3^a + 7^b which result in a perfect square answer.
I'm not sure where to begin. I have an idea that there are several
situations to analyze. For example if a and be are equal they can be
factored by a formula and if they are both even they are going to be
complex roots.

Date: 06/29/2009 at 08:36:01
From: Doctor Jacques
Subject: Re: find ordered pair (a,b) for 3^a + 7^b giving a perfect square
Hi Dan,
This is a very nice exercise about congruences. Although the proof
does not require any elaborate concept, it requires a significant
amount of work--be prepared for a hard ride...
We are looking for solutions of:
3^a + 7^b = n^2 [1]
where a, b, and n are non-negative integers.
We can first look at [1] as a congruence modulo 7:
3^a = n^2 (mod 7) [2]
Modulo 7, the quadratic residues (i.e., the squares) are 0, 1, 2, and
4, and 3 is not one of them. And we also have
3^6 = 1 (mod 7) [3]
by Fermat's theorem. This implies that a must be even.
[We could also look at congruences modulo 4 to deduce that b must be
odd, but we will not need that fact.]
As a is even, we may write a = 2k for some integer k. We have:
3^(2k) + 7^b = n^2
n^2 - 3^(2k) = 7^b
(n + 3^k)(n - 3^k) = 7^b [4]
This means that the product on the LHS must be a power of 7, and
therefore each factor must also be a power of 7.
However, it is not possible to have both factors on the LHS divisible
by 7, as this would imply that their difference
(n + 3^k) - (n - 3^k) = 2*3^k
would be divisible by 7, which is clearly impossible. The only power
of 7 that is not a multiple of 7 is 7^0 = 1. We may therefore
conclude that one of the factors must be 1, and this is obviously the
smaller factor; we have:
n - 3^k = 1
n = 3^k + 1 [5]
If we substitute that into [4], we obtain:
2*3^k + 1 = 7^b
2*3^k = 7^b - 1 [6]
and this means that 7^b - 1 must be twice a power of 3.
Let us look at the simple cases first. If k = 0, we have 7^b - 1 = 2,
which is impossible. If k = 1, we have:
6 = 7^b - 1
which has the solution b = 1. We will now show that this is the only
solution. We may now assume that k > 1, and [6] shows that 7^b - 1
must be divisible by 9; in other words, we must have:
7^b = 1 (mod 9) [7]
Modulo 9, the powers of 7 are:
m | 0 1 2 3
----+--------
7^m | 1 7 4 1
and this shows that [7] will be satisfied if and only if b is a
multiple of 3. We have:
7^3 - 1 = 342 = 2*3^2*19
but this means that:
7^3 = 1 (mod 19)
and we will have 7^b = 1 (mod 19) for all exponents b that are
multiples of 3; this is the same as saying that, if b is a multiple of
3, 7^b - 1 is a multiple of 19, which contradicts equation [6].
Therefore, the only solution is k = 1, a = 2 and b = 1, corresponding
to 3^2 + 7^1 = 4^2.
Please feel free to write back if you require further assistance.
- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/